1999 JBMO Problems/Problem 4

Problem 4

Let $ABC$ be a triangle with $AB=AC$ . Also, let $D\in[BC]$ be a point such that $BC>BD>DC>0$ , and let $\mathcal{C}_1,\mathcal{C}_2$ be the circumcircles of the triangles $ABD$ and $ADC$ respectively. Let $BB'$ and $CC'$ be diameters in the two circles, and let $M$ be the midpoint of $B'C'$ . Prove that the area of the triangle $MBC$ is constant (i.e. it does not depend on the choice of the point $D$ ).

Solution

Its easy to see that $B'$ , $C'$ , $D$ are collinear (since $\angle B'DC = \angle C'DC$ = 90^\circ $). Applying the sine rule in triangle$ ABC $, we get$ \frac{\sin BAD }{ \sin CAD} = \frac{BD }{ DC}. $Since$ BAB'D $and$ CC'AD $are cyclic quadrilaterals,$ \angle BAD \angle BB'D $and$ \angle CAD = \angle CC'D. $So,$ \frac{\sin(BB'D)}{\sin(CC'D)} = \frac{BD}{DC} $and$ \frac{BD}{\sin BB'D} = \frac{DC }{ \sin CC'D}. $Thus,$ BB' = CC' $(the circumcircles$ \mathcal{C}_1,\mathcal{C}_2$are congruent).

From right triangles$ (Error compiling LaTeX. Unknown error_msg)BB'A $and$ CC'A $, we have <cmath>AC'^{2} = CC'^{2} - AC^{2} = BB'^{2} - AB^{2} = AB'^{2}</cmath> So$ AC' = AB'. $Since$ M $is the midpoint of$ B'C' $,$ AM $is perpendicular to$ B'C' $and hence$ AM $is parallel to$ BC $. So area of$ [MBC] = [ABC] $and hence is independent of position of$ D $on$ BC$.

Page

Toolbox

Search

1999 JBMO Problems/Problem 4

Problem 4

Solution